MATHEMATICIANS..Please help, guys!!!!?

Question

Hi, Below is a question of polynomials. Could you please help me get to know how the answer is obtained in a detailed way? Please make sure u explain the answer and its steps clearly so that i can understand it. Thanks a lot in advance!!

Qn: When a polynomial, P(x), is divided by x-a, it leaves a remainder of a^3 and when it is divided by x-b, it leaves a remainder of b^3. Find the remainder when P(x) is divided by (x-a)(x-b).

Note -> The answer to this question is (a^2 + ab + b^2)x - ab(a + b). However, I dont have a single idea as to how u get this answer. So please explain clearly. VERY CLEARLY.

Thanks.

Answer 1

Let,
P(x) = (x - a) * Q1(x) + a^3 , P(a) = a^3
P(x) = (x - b) * Q2(x) + b^3 , P(b) = b^3

P(x) = (x - a)(x - b) * Q3(x) + R(x) ......................(1), here remainder is R(x), which is 1st degree polynomial as (x - a)(x - b) is 2nd degree polynomial.

If x = a in eq. (1),
P(a) = R(a)
or,R(a) = a^3
that means when R(x) is divided by (x - a), the remainder will be a^3.

Therefore,
R(x) = R1(x - a) + a^3 ...................................... (2), here R1 is a constant because R(x) is a 1st degree polynomial.

If x = b in eq. (1),
P(b) = R(b)
or,R(b) = b^3
that means when R(x) is divided by (x - b), the remainder will be b^3.

Therefore,
R(x) = R2(x - b) + b^3 ........................................ (3)

Comparing eqns. (2) and (3),
R1(x - a) + a^3 = R2(x - b) + b^3 ................................ (4)

Equating the coefficients of x,
R1 = R2 = R (say)

Then, eq (4) becomes,
Rx - Ra + a^3 = Rx - Rb + b^3
or,-Ra + a^3 = -Rb + b^3
or,R(a - b) = a^3 - b^3
or,R = a^2 + ab + b^2

Putting this value in eq. (2) or (3),

R(x) = (a^2 + ab + b^2) (x - a) + a^3
or,R(x) = (a^2 + ab + b^2)x - (a^3 + a^2b + ab^2) + a^3
or, R(x) = (a^2 + ab + b^2)x - ab(a + b)

Thus, the remainder is (a^2 + ab + b^2)x - ab(a + b).

Answer 2

Here goes the answer

Let P(x) be the polynomial P(x) = L1+L2x+L3 x^2+L4 x^3+.......

since a^3 is a remainder with x-a therefore from the remainder theorem P(a) = a^3, which is only possible when all the constants L1, L2, ........ except L4 are 0, L4 being equal to 1

so, P(x) = a^3 or x^3

because of the other case with x-b we can say that
P(x) = x^3

now dividing x^3 with (x-a)(x-b) which can be easily done
gives remainder = (a^2 + ab + b^2)x - ab(a + b) and

quotient = x+(a+b)

I hope its clear

Answer 3

According to the properties of polynomials (I assume you aware of them) we have P(a) = a^3 and P(b) = b^3.

The quotient of the division of P by (x-a)(x-b)is a polynomial Q and the remainder is a polynomial of the first degree, that is, a polynomial R given by R(x) = px + q. Then, P(x) = Q(x)(x-a)(x-b) + R(x). Making x=a and x = b in this equation, we get R(a) = a^3 and R(b) = b^3. I'm assuming a<> b. Now, all we have to do is solve for p and q the system of linear equations

p*a + q = a^3
p*b + q = b^3.

If we subtract these equations we get p(a- b) = a^3 - b^3 and, since a<>b, it follows that p = (a^3 - b^3)/(a -b) = a^2 + ab + b^2. If we plug this expression of p in one of the 2 equations and solve for q, we get q = -ab(a+ b). Thefore your remainder is R(x) = (a^2 + ab + b^2)x - ab(a+b) .

Next time, try to do your homework, OK?.

Answer 4

You're question is going to need a considerable amount of calculations.

The answer to the first two equations is the variable substracted from x cubed. So x-a becomes a^3 and x-b becomes b^3.

In the question (x-a)(x-b) the answer is ((x-a)(x-b))^3. That is,
(x-a)(x-b) * (x-a)(x-b) * (x-a)(x-b).

This becomes (x^2-bx-ax+ab)^3.

So, unless someone has a better solution, you will need to multiply (x^2-bx-ax+ab) * (x^2-bx-ax+ab) * (x^2-bx-ax+ab)

That is, you multiply everything in the first equation by x^2 then by -bx etc.

You then take that number and multiply it by x^2, then -bx etc again.

Finally, you factor out all the common elements.

i realise this sounds like a ton of work and there may be a better way to do this but this is the only method I know. Good luck.

Answer 5

let R be the remainder.
as R is a remainder of a polynomial we assume
R = Ax + B
By question
R(a) = Aa + B = a^3 .................(i) &
R(b) = Ab + B = b^3 ................(ii)
Subtracting (ii) from (i)
A (a-b) = a^3-b^3
=> A = (a^3-b^3)/(a-b)
=> A = a^2 + ab + b^2 ( considering a≠b)
Putting this value of a in Eq. (i)
B = a^3 - a(a^2 + ab + b^2) = a(a^2 - a^2 -ab - b^2)
= -ab(a+b)

=> remainder = (a^2 + ab + b^2)x - ab(a+b)

Answer 6

Go to www.algebra.com it is special website to help with math.

Answer 7

Lol they should have a Yahoo! Homeworks..

Answer 8

Thanks for the question. This was bugging me too.